Answer
(a) $10.6m$
(b) less than $3.0s$
Work Step by Step
(a) We can determine the depth of the well as follows:
$d=ut+\frac{1}{2}gt^2$....eq(1)
$\implies d=0(t)+\frac{1}{2}(9.8m/s^2)t^2$
$d=(4.9m/s^2)t^2$.....eq(2)
Time required by sound to reach up the well
$t_1=1.5s-t$
Now eq(1) becomes
$d=340m/s(1.5s-t)+\frac{1}{2}(0m/s^2)t^2$
$\implies d=510m-(340m/s)t$...eq(3)
comparing eq(2) and eq(3), we obtain:
$(4.9m/s^2)t^2=510m-(340m/s)t$
This simplifies to:
$t=1.47s$
Now we plug in this value in eq(2) to obtain:
$d=(4.9m/s^2)(1.47s)^2$
$\implies d=10.6m$
(b) We know that if the depth of the well is doubled then
$d=ut_2+\frac{1}{2}gt_2^2$
We plug in the known values to obtain:
$20.4m=(0m/s)t_2+\frac{1}{2}(9.8m/s^2)t_2^2$
This simplifies to:
$t_2=2.04s$
Now $v=\frac{d}{t_3}$
$340m/s=\frac{20.4m}{t_3}$
$\implies t_3=0.06s$
Thus, the time taken to hear the splash is given as
$T=t_2+t_3$
$\implies T=2.04s+0.06s$
$\implies T=2.1s$
We conclude that, if the depth of the well is doubled then the sound of splash is heard in less than $3.0s$.
